general Nakayama's lemma

Let \(m_1,...,m_n\) be generators of \(M\).
Then by assumption, as \(\mathfrak{a}M = M\), each \(m_i\) can be written as \(m_i = \sum_{j=1}^{n} \alpha_{i,j} m_j\) with \(\alpha_{i,j} \in \mathfrak{a}\).
Therefore also

\begin{align*} 0 =& m_i - m_i \\ =& \sum \alpha_{i,j} m_j - \delta_{i,j} m_i \end{align*}

where \(\delta_i\) is the Kronecker-Delta.

Let \(B\) be the matrix

\begin{align*} B =& (\delta_{i,j} - \alpha_{i,j}) \\ =& \begin{pNiceMatrix} 1 - \alpha_{1,1} & - \alpha_{1,2} & ... & - \alpha_{1,n} \\ - \alpha_{2,1} & - \alpha_{2,2} & \ddots & -\alpha_{2,n} \\ \vdots & \ddots & \ddots & \vdots \\ - \alpha_{n,1} & - \alpha_{n,2} & ... & 1 - \alpha_{n,n} \\ \end{pNiceMatrix} \end{align*}

and \(d \coloneqq \mathrm{det}(B)\).

Then

\begin{align*} d = 1 - a \end{align*}

for some \(a \in \mathfrak{a}\) since it is given by the ones in the diagonal and a linear combination where at least one factor is an element of \(\mathfrak{a}\)

It remains to show that \(d M= 0\).
It suffices to show that \(d \cdot m_h =0\) for a generator \(m_h\).

Let \(\tilde{B} = (\tilde{b}_{i,j})\) be the Adjunkte with

\begin{align*} \tilde{B} \cdot B = d \cdot E_n \end{align*}

(cf. Multiplikation einer Matrix mit der Adjunkte )

Then the entry in the \(h,j\)-spot is given by \(\sum_{i=1}^n \tilde{b}_{h,i} b_{i,j} = d \delta_{h,j}\)

This shows

\begin{align*} d m_h =& d E_n(m_h) \\ =& d \tilde{B} B m_h \\ =& \sum_{j=1}^n d \partial_{h,j} m_j \\ =& \sum_{j=1}^n \sum_{i=1}^n \tilde{b}_{h,i} \cdot b_{i,j} m_j \\ =& \sum_{i=1}^n \sum_{j=1}^n b_{i,j} m_j \tilde{b}_{h,i} \end{align*}

Since \(\sum_{j=1}^n b_{i,j} m_j = \sum_{j=1}^n (\partial_{i,j} - \alpha_{i,j}) m_j = 0\) this shows that

\begin{align*} d m_h = 0 \end{align*}

and thus we are done.